Electronic – a typical diode’s ideality factor

Schottky diodes do not have reverse recovery time. Recovery from what? In a normal p-n junction diode, there is a charge carrier depletion region, and so the correct polarity electric field applied (the voltage drop) is actually switching it from non-conducting to conducting. If that field is removed, or applied in the opposite polarity, it is switched off again, but p-n junction diodes are very much switches that must turn on and off, and take time to do so, and that is the recovery time.

Schottky diodes are not constructed using two semiconductor junctions like p-n diodes. They are a metal-semiconductor junction. Due to some pretty nontrivial quantum physics which is beyond the scope of this question, Schottky diode junctions actually behave like true one-way valves. Something called the work function, which is the energy needed to 'dislodge' an electron out of a material and into the vacuum directly adjacent to the material, is very high for metals, but very low for semiconductors, at least when they form a junction with each other. Again, this is a huge oversimplification, and there are a lot of other things going on, but the gist is that the interface of the metal and semiconductor create a very tiny 'vacuum' depletion zone, one that is easily crossed via thermionic emission (yes, like how a vacuum tube works) from the semiconductor to the metal, because the work function is very low in the semiconductor. But in the metal, the work function is very high indeed, and it just takes too much energy to dislodge electrons out of the metal and into the semiconductor. A few electrons do make it, but because they are statistical outliers that managed to get the huge amount of thermal excitation needed to leave the metal. Otherwise, electrons pass easily from semiconductor to metal, but pass essentially not at all from the metal to semiconductor.

So, Schottky diodes do not have reverse recovery time because they do not have anything to recover from. However, the vacuum is effectively acting as a dielectric in one direction, so there is some small amount of parasitic capacitance. The reverse current seen in Schottky diodes is not actually reverse conduction, but merely a capacitive discharge. This is why Schottky's are said to have 'soft' recovery, as the curve is really just a capacitor discharge curve, and that takes time. But it is not 'on' and allowing reverse current flow. All the current flowing in reverse is due to energy stored capacitively from the diode itself.

One final caveat: In the larger, high power Schottky diodes, due to their physical construction (to shape the electric field so as to not cause dielectric breakdown across the vacuum barrier) have a guard ring that forms an entirely separate parasitic p-n junction in the Schottky diode. With low forward bias, it remains largely invisible, and the capacitance is all that matters. This is why datasheets always have the reverse recovery time listed for a very small forward voltage. Unfortunately, as the forward bias increases, it will eventually turn on the parasitic p-n diode junction through which reverse current can flow until switched off, thus vastly increasing the effective recovery time. The Schottky junction itself is still without a recovery time, as it has nothing to recover from, but the separate parasitic p-n junction does need to recover.

So be warned, the reverse recovery times for high power Schottky diodes are generally measured with forward bias too low to turn on this parasitic junction, but in real world applications, the recovery time mentioned is, and this is being generous, "very optimistic." It's frustrating (and intentional) that the recovery times under higher biases are often left out entirely of datasheets.

No, the ideality factor should be between 1 and 2 inclusive. Of course with some random set of data it might not be representative of a real diode.

However, I don't think your optimization is very good. Below is the same exercise which I did myself. The ideality factor comes out well within the above range and Is at a quite reasonable 1.16pA. As you can see, the two curves are pretty much on top of each other.

Since this looks like a homework question, that should be enough of a hint.

Okay, using MATLAB, here is a similar fit (I used amperes for the Y axis). The value of Is came out a bit different at 1.243pA (I would trust this one more) and the ideality factor very similar to Excel. Again, as you can see, the data points are right on top of the fit. If I use the bisquare or robust algorithms, Is comes out very similar to the Excel version (1.17pA vs. 1.16pA).